11.45.1 The ordinary model and the greatest solutions27CD354DC77F

Proof. First take the ordinary envelope on the curve with background $K\omega _1$ and obstacle $\min (\psi +C,0)$. Rotation of $z$ preserves both the obstacle and the admissible family, so the envelope is rotation invariant. With $s=\log |z|^2$ and $F(s)=K\log (1+\mathrm{e}^ s)$, its profile plus $F$ is convex and lies below $F$. The comparison with $\psi +C$ bounds its final slope by $a$. At $s_*=\log R$, one has $F'(s_*)=a$. Above $s_*$, convexity and the final slope bound give

and below $s_*$ the obstacle gives $u\leq 0$. The function in (11.45.2) attains both bounds. Its convex profile, including its logarithmic value at infinity, gives a curve potential, and direct calculation gives $\psi \leq \phi _{a,d}\leq \psi +C_*$. It is therefore a competitor for every $C\geq C_*$, so the curve envelope is exactly $\phi _{a,d}$.

A product competitor restricts to each curve $\mathbb P^1\times \{ y\} $ as either the constant $-\infty $ or a curve competitor. It is thus bounded above by $\phi _{a,d}$. Conversely, the curve model is a product competitor because the vertical form $b\kappa $ is positive. Hence the product envelope is the same function. Ordinary-type invariance of the singularity envelope gives its model property. Moreover,

which proves the smaller-background assertion.

Off $D$, the horizontal curvature of $\phi _{a,d}$ is $K\mathbf1_{\{ |z|^2<R\} }\omega _1$. The first derivatives of its convex profile agree at $R$, so there is no boundary-circle atom. Expanding the product off $D$ gives the displayed measure, and the non-pluripolar product assigns $D$ zero mass. Finally, in terms of $\ell $ the residual extension is

The formulas agree at the moving interface and extend to both endpoints. On compact positive parameter ranges each logarithm is used away from its singular endpoint. Joint continuity on the resulting compact set proves uniform variation. $\square$

Proof. The one-sided ordinary class has the description

Indeed, $u\preceq \psi $ gives $\nu (u,D)\geq d$. Divisorial removal, lemma 5.38.1, shows that $\theta +\mathrm{dd}^{\mathrm{c}}u-d[D]\geq 0$. Thus the locally integrable difference $u-\psi $ has an $\eta $-plurisubharmonic representative $r$. The equality $u=\psi +r$ holds off $D$ by canonical equality and on $D$ because both sides equal $-\infty $. Conversely, adding $\psi $ to $r$ adds the positive current $d[D]$; since $r$ has a finite compact upper bound, the one-sided comparison follows. Off $D$ the currents agree, so finite-level locality identifies their extended products, including at any additional poles of $r$. Both measures give $D$ zero mass, proving the global equality. The bounded ordinary type corresponds exactly to bounded $r$, as in theorem 11.41.1.

The function $g_\sigma =\mathrm{e}^{\sigma \psi }f$ extends continuously by zero on $D$ and is positive elsewhere. Since $\eta >0$, theorem 14.19.3 gives a continuous solution $R_\sigma $ of

For this bounded density, the greatest solution of theorem 15.35.2 is bounded and coincides with $R_\sigma $ by corollary 14.19.12. Thus it dominates every $\eta $-plurisubharmonic exponential subsolution. The correspondence (11.45.6) transfers this order to the entire one-sided ordinary class and gives $s_\sigma =\psi +R_\sigma $. Bounded residual uniqueness gives uniqueness in the ordinary type. The bounded residual equation has finite mass, so the singular equation does too. $\square$